All categories

3 years ago

A point charge has q=1.0×10-6

Physics

1 answer:

dmitriy555 [2]3 years ago

8 0

Since the 2 points form a triangle with hypothenuse of √5 [ √(1²+2²)],

I guess apply the formula :

$v = kq \div r$

with r as √5 and q as 1x10^-6

not sure about this answer tho

draw a diagram first to understand better

You might be interested in

If 2.2 lbs = 1.0 kg, and Megan Progress weighs 130 lbs, what is her weight in newtons? W = N (Round your answer to nearest whole

lord [1]

as it is given that

2.2 lbs = 1 kg

here we know that Megan Progress weighs 130 lbs

so its mass in kg is given as

$m = 130 * \frac{1}{2.2} = 59.1 kg$

now to find the weight in Newton unit we can say

$W = mg$

$W = 59.1* 9.8$

$W = 579 N$

so 130 lbs weighs as 579 N in SI units

5 0

3 years ago

You have a light spring which obeys Hooke's law. This spring stretches 2.92 cm vertically when a 2.70 kg object is suspended fro

ehidna [41]

(a) 907.5 N/m

The force applied to the spring is equal to the weight of the object suspended on it, so:

$F=mg=(2.70 kg)(9.8 m/s^2)=26.5 N$

The spring obeys Hook's law:

$F=k\Delta x$

where k is the spring constant and $\Delta x$ is the stretching of the spring. Since we know $\Delta x=2.92 cm=0.0292 m$ , we can re-arrange the equation to find the spring constant:

$k=\frac{F}{\Delta x}=\frac{26.5 N}{0.0292 m}=907.5 N/m$

(b) 1.45 cm

In this second case, the force applied to the spring will be different, since the weight of the new object is different:

$F=mg=(1.35 kg)(9.8 m/s^2)=13.2 N$

So, by applying Hook's law again, we can find the new stretching of the spring (using the value of the spring constant that we found in the previous part):

$\Delta x=\frac{F}{k}=\frac{13.2 N}{907.5 N/m}=0.0145 m=1.45 cm$

(c) 3.5 J

The amount of work that must be done to stretch the string by a distance $\Delta x$ is equal to the elastic potential energy stored by the spring, given by:

$W=U=\frac{1}{2}k\Delta x^2$

Substituting k=907.5 N/m and $\Delta x=8.80 cm=0.088 m$ , we find the amount of work that must be done:

$W=\frac{1}{2}(907.5 N/m)(0.088 m)^2=3.5 J$

5 0

3 years ago

During a race, a runner runs at a speed of 6 m/s. 2 seconds later, she is running at a speed of 10 m/s. What is the runner’s acc

Lina20 [59]

V o = 6 m/s,
t = 2 s
v = 10 m/s
v = v o + a t
a t = v - v o
a = ( v - v o ) / t
a = ( 10 m/s - 6 m/s ) / 2 s = 4 m/s / 2 s = 2 m/s²
Answer:
The runner`s acceleration is 2 m/s².

6 0

3 years ago

A speed skater is sliding across the ice at a speed of 19 m/s when she encounters a rough patch of ice that is 2.0 m wide. As sh

Stells [14]

Answer:

$-9.25 m/s^2$

Explanation:

The motion of the skater is a uniformly accelerated motion, so we can find the deceleration by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

In this problem, we have:

u = 19 m/s is the initial velocity of the skater

v = 18 m/s is the final velocity of the skater

s = 2.0 m is the displacement

Solving for a, we can find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{18^2-19^2}{2(2.0)}=-9.25 m/s^2$

And the negative sign tells us that the acceleration is in the opposite direction to the motion (so, it is a deceleration)

5 0

4 years ago

The strength of an electromagnet can be altered by

denpristay [2]

The strength of an electromagnet can be altered by increasing the number of coils around the core. The more times the coil is wrapped, the stronger the electromagnet is.

Your answer is: B) Increasing the number of coils around the core

Have an amazing day and stay hopeful!

3 0

3 years ago

Read 2 more answers